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 Approximation method for super square root Ztolk Junior Fellow Posts: 21 Threads: 4 Joined: Mar 2010 03/23/2010, 01:09 AM (This post was last modified: 03/23/2010, 01:10 AM by Ztolk.) I came up with this method for calculating x where x^x=y. It's similar to a method for calculating square roots. The procedure is as follows: 1. Pick the number y that you would like to find the super root of. 2. Find the self-power that is closest to and less than y. We call this t^t. For example, if y=4000 then t=5 and t^t=3125. 3. Calculate the difference between y and t^t. So for y=4000, t=5, we get 875. 4. Divide that difference by the interval between the self-power below y and the one above it. So for y=4000, this would be (6^6-5^5)=46656-3125=43531. 5. Add this to t. So with our example the result is 5+875/43531 which is about 5.0201. The actual value is 5.094, so the approximation is 1.5% off the actual value. The approximation generally gets more accurate as y and t get higher. Within the regime of a specific t, the most inaccurate value will occur when y is about 27.4% through the interval (example, for t=5, the approximation is most inaccurate at y=11927+3125=15092). I haven't figured out why this is. The accuracy at this point decreases with increasing t. The integer for which this yields the least accurate value is 2, at about 85%. This method is much more accurate than a first order Taylor approximation of the actual solution. I haven't yet been able to apply this to higher order tetrations. « Next Oldest | Next Newest »

 Messages In This Thread Approximation method for super square root - by Ztolk - 03/23/2010, 01:09 AM RE: Approximation method for super square root - by bo198214 - 03/23/2010, 10:54 AM RE: Approximation method for super square root - by Ztolk - 03/23/2010, 02:33 PM

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